Two dimensional collisions Inelastic: How to find magnitude of velocity and angl

Question

Two dimensional collisions Inelastic: How to find magnitude of velocity and angle?

x compenent: (2.65e-26kg *350m/s)-(3.0e-26kg*410 m/s) = (5.66e-26) (V final * cos[theta] )

-53.54 = Vf (Cos[theta])

Y compenent: (2.65e-26 kg*-80m/s)+(3.0e-26 kg* 105 m/s) = (5.66e-26) (V final* sin[theta])

17.9= Vf sin[theta]

The answers ended up being = 55.7 m/s and 162 degrees or 2.83 Radions

I'm confused as to how he got htose answers, I udnerstand up until the problems I wrote out...what equation did he use to find the angle ?

The question that corresponds with this is:

Methane (2.65e -26 kg) collides inelastically with water (3.0e -26 kg). The initial velocities and speed are methane -> 350x-80y m/s while for water -410x+105 y m/s.

Explanation / Answer

let

-53.54 = Vf*cos(theta) ---(1)

17.9 = Vf*sin(theta) ---(2)

take equation(2)/equation(1)

17.9/(-53.54) = Vf*sin(theta)/(Vf*cos(theta))

-0.3343 = tan(theta)

==> theta = tan^-1(-0.3343)

= 18 degrees with -x axis in clockwise direction.

so, 180 - 18 = 162 degrees with +x axis in counter clocwise direction.

now from equation 1

-53.54 = Vf*cos(162)

==> Vf = -53.54/cos(162)

= 56.3 m/s <<<<<<<-------Answer

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.